Mixing
Basis functions and positive definiteness
$begingroup$
Suppose I have a set of functions $g_i(x)$ which form a basis on some interval $[a,b]$. My question is whether the matrix
$$
A_{ij} = int_a^b g_i(x) g_j(x) dx
$$
is positive definite?
My approach so far is to consider a function $f: [a,b] rightarrow mathbb{R}$,
$$
f(x) = sum_i c_i g_i(x) = c^T g(x)
$$
Then we have:
begin{align}
0 &le int_a^b |f(x)|^2 dx \
& = int_a^b c^T g(x) g(x)^T c dx \
&= c^T left( int_a^b g(x) g(x)^T dx right) c \
&= c^T A c
end{align}
This shows that $A$ is postive semi-definite.
But I wonder if it is correct to argue that $int_a^b |f(x)|^2 dx$ is equal to $0$ only if $c = 0$, because otherwise if $c ne 0$, then that implies that the $g_i(x) = 0$ for all $i$ and all $x$, contradicting the assumption that $g_i(x)$ is a basis. Is that correct reasoning?
As a followup question, consider the matrix
$$
B_{ij} = int_a^b g'_i(x) g'_j(x) dx
$$
This matrix is positive semi-definite by the same reasoning above, but is it also positive definite? If the argument above is correct, I don't think it applies to the matrix $B$, because if $g_i(x)$ forms a basis on $[a,b]$, that doesn't necessarily mean that $g'_i(x)$ is also a basis?
So the question is whether $A$ and/or $B$ are positive definite matrices?
linear-algebra matrices
asked Jan 18 at 7:12
vibevibe
1648
$endgroup$
add a comment |
$begingroup$
Suppose I have a set of functions $g_i(x)$ which form a basis on some interval $[a,b]$. My question is whether the matrix
$$
A_{ij} = int_a^b g_i(x) g_j(x) dx
$$
is positive definite?
My approach so far is to consider a function $f: [a,b] rightarrow mathbb{R}$,
$$
f(x) = sum_i c_i g_i(x) = c^T g(x)
$$
Then we have:
begin{align}
0 &le int_a^b |f(x)|^2 dx \
& = int_a^b c^T g(x) g(x)^T c dx \
&= c^T left( int_a^b g(x) g(x)^T dx right) c \
&= c^T A c
end{align}
This shows that $A$ is postive semi-definite.
But I wonder if it is correct to argue that $int_a^b |f(x)|^2 dx$ is equal to $0$ only if $c = 0$, because otherwise if $c ne 0$, then that implies that the $g_i(x) = 0$ for all $i$ and all $x$, contradicting the assumption that $g_i(x)$ is a basis. Is that correct reasoning?
As a followup question, consider the matrix
$$
B_{ij} = int_a^b g'_i(x) g'_j(x) dx
$$
This matrix is positive semi-definite by the same reasoning above, but is it also positive definite? If the argument above is correct, I don't think it applies to the matrix $B$, because if $g_i(x)$ forms a basis on $[a,b]$, that doesn't necessarily mean that $g'_i(x)$ is also a basis?
So the question is whether $A$ and/or $B$ are positive definite matrices?
linear-algebra matrices
asked Jan 18 at 7:12
vibevibe
1648
$endgroup$
$begingroup$
I have edited my answer and included a counterexample for $B$.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 12:47
add a comment |
$begingroup$
Suppose I have a set of functions $g_i(x)$ which form a basis on some interval $[a,b]$. My question is whether the matrix
$$
A_{ij} = int_a^b g_i(x) g_j(x) dx
$$
is positive definite?
My approach so far is to consider a function $f: [a,b] rightarrow mathbb{R}$,
$$
f(x) = sum_i c_i g_i(x) = c^T g(x)
$$
Then we have:
begin{align}
0 &le int_a^b |f(x)|^2 dx \
& = int_a^b c^T g(x) g(x)^T c dx \
&= c^T left( int_a^b g(x) g(x)^T dx right) c \
&= c^T A c
end{align}
This shows that $A$ is postive semi-definite.
But I wonder if it is correct to argue that $int_a^b |f(x)|^2 dx$ is equal to $0$ only if $c = 0$, because otherwise if $c ne 0$, then that implies that the $g_i(x) = 0$ for all $i$ and all $x$, contradicting the assumption that $g_i(x)$ is a basis. Is that correct reasoning?
As a followup question, consider the matrix
$$
B_{ij} = int_a^b g'_i(x) g'_j(x) dx
$$
This matrix is positive semi-definite by the same reasoning above, but is it also positive definite? If the argument above is correct, I don't think it applies to the matrix $B$, because if $g_i(x)$ forms a basis on $[a,b]$, that doesn't necessarily mean that $g'_i(x)$ is also a basis?
So the question is whether $A$ and/or $B$ are positive definite matrices?
linear-algebra matrices
asked Jan 18 at 7:12
vibevibe
1648
$endgroup$
Suppose I have a set of functions $g_i(x)$ which form a basis on some interval $[a,b]$. My question is whether the matrix
$$
A_{ij} = int_a^b g_i(x) g_j(x) dx
$$
is positive definite?
My approach so far is to consider a function $f: [a,b] rightarrow mathbb{R}$,
$$
f(x) = sum_i c_i g_i(x) = c^T g(x)
$$
Then we have:
begin{align}
0 &le int_a^b |f(x)|^2 dx \
& = int_a^b c^T g(x) g(x)^T c dx \
&= c^T left( int_a^b g(x) g(x)^T dx right) c \
&= c^T A c
end{align}
This shows that $A$ is postive semi-definite.
But I wonder if it is correct to argue that $int_a^b |f(x)|^2 dx$ is equal to $0$ only if $c = 0$, because otherwise if $c ne 0$, then that implies that the $g_i(x) = 0$ for all $i$ and all $x$, contradicting the assumption that $g_i(x)$ is a basis. Is that correct reasoning?
As a followup question, consider the matrix
$$
B_{ij} = int_a^b g'_i(x) g'_j(x) dx
$$
This matrix is positive semi-definite by the same reasoning above, but is it also positive definite? If the argument above is correct, I don't think it applies to the matrix $B$, because if $g_i(x)$ forms a basis on $[a,b]$, that doesn't necessarily mean that $g'_i(x)$ is also a basis?
So the question is whether $A$ and/or $B$ are positive definite matrices?
linear-algebra matrices
linear-algebra matrices
asked Jan 18 at 7:12
vibevibe
1648
asked Jan 18 at 7:12
vibevibe
1648
asked Jan 18 at 7:12
vibevibe
1648
asked Jan 18 at 7:12
vibevibe
1648
asked Jan 18 at 7:12
vibevibe
1648
1648
$begingroup$
I have edited my answer and included a counterexample for $B$.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 12:47
add a comment |
$begingroup$
I have edited my answer and included a counterexample for $B$.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 12:47
$begingroup$
I have edited my answer and included a counterexample for $B$.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 12:47
$begingroup$
I have edited my answer and included a counterexample for $B$.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 12:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b}sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b} |sum c_ig_i'(x)|^{2}, dx geq 0$. So $B$ is non-negative definite. But $sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=0$ only implies $sum c_ig_i=c$ for some constant $c$ and $B$ need not be positive definite. For a counterexample consider $g_1(x)=x,g_2(x)=1-x$ on $[0,1]$. It is easy to see that $B$ is not positive definite.
answered Jan 18 at 7:16
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
Thank you, this makes sense
$endgroup$
– vibe
Jan 18 at 15:37
add a comment |
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$begingroup$
$sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b}sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b} |sum c_ig_i'(x)|^{2}, dx geq 0$. So $B$ is non-negative definite. But $sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=0$ only implies $sum c_ig_i=c$ for some constant $c$ and $B$ need not be positive definite. For a counterexample consider $g_1(x)=x,g_2(x)=1-x$ on $[0,1]$. It is easy to see that $B$ is not positive definite.
answered Jan 18 at 7:16
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
Thank you, this makes sense
$endgroup$
– vibe
Jan 18 at 15:37
add a comment |
$begingroup$
$sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b}sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b} |sum c_ig_i'(x)|^{2}, dx geq 0$. So $B$ is non-negative definite. But $sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=0$ only implies $sum c_ig_i=c$ for some constant $c$ and $B$ need not be positive definite. For a counterexample consider $g_1(x)=x,g_2(x)=1-x$ on $[0,1]$. It is easy to see that $B$ is not positive definite.
answered Jan 18 at 7:16
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
Thank you, this makes sense
$endgroup$
– vibe
Jan 18 at 15:37
add a comment |
$begingroup$
$sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b}sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b} |sum c_ig_i'(x)|^{2}, dx geq 0$. So $B$ is non-negative definite. But $sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=0$ only implies $sum c_ig_i=c$ for some constant $c$ and $B$ need not be positive definite. For a counterexample consider $g_1(x)=x,g_2(x)=1-x$ on $[0,1]$. It is easy to see that $B$ is not positive definite.
answered Jan 18 at 7:16
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b}sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=int_a^{b} |sum c_ig_i'(x)|^{2}, dx geq 0$. So $B$ is non-negative definite. But $sum c_ic_j int_a^{b} g_i'(x)g_j'(x), dx=0$ only implies $sum c_ig_i=c$ for some constant $c$ and $B$ need not be positive definite. For a counterexample consider $g_1(x)=x,g_2(x)=1-x$ on $[0,1]$. It is easy to see that $B$ is not positive definite.
answered Jan 18 at 7:16
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
edited Jan 18 at 12:46
answered Jan 18 at 7:16
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Jan 18 at 7:16
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Jan 18 at 7:16
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
62.9k42362
$begingroup$
Thank you, this makes sense
$endgroup$
– vibe
Jan 18 at 15:37
add a comment |
$begingroup$
Thank you, this makes sense
$endgroup$
– vibe
Jan 18 at 15:37
$begingroup$
Thank you, this makes sense
$endgroup$
– vibe
Jan 18 at 15:37
$begingroup$
Thank you, this makes sense
$endgroup$
– vibe
Jan 18 at 15:37
add a comment |
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$begingroup$
I have edited my answer and included a counterexample for $B$.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 12:47